Evaluate $\lim_{x \to 0^-} \left( \frac{1}{x} - \frac{1}{|x|} \right)$ (possible textbook mistake - James Stewart 7th)

Question

I was working on a few problems from James Stewart's Calculus book (seventh edition) and I found the following:

Find

$$\lim_{x \to 0^-} \left( \frac{1}{x} - \frac{1}{|x|} \right)$$

Since there's a $|x|$ on the limit and knowing that $|x| = -x$ for any value less than zero, we have

$$\lim_{x \to 0^-} \left( \frac{1}{x} - \frac{1}{|x|} \right) = \lim_{x \to 0^-} \frac{2}{x}$$

So far so good. Continuing,

$$\lim_{x \to 0^-} \left( \frac{1}{x} - \frac{1}{|x|} \right) = \lim_{x \to 0^-} \frac{2}{x} = - \infty$$

since the denominator becomes smaller and smaller. When checking the textbook's answer I've found the following:

Am I missing something or should the limit really be $- \infty$ ?

Answer 1

Saying that the limit is equal to $\infty$ is a mathematical shorthand (amongst some mathematicians, at least) for:

Given any real number $M$, there is a real $\delta$ (depending on $M$) such that $\frac{1}{x^2} > M $ for all $x$ satisfying $0 <x <|\delta|$.

It is usually advised that beginners avoid using $\infty$ since it leads to careless or wrong manipulations of the symbol all too often.

Answer 2

bru, I think that the problem is just about terminology. Your derivation is correct, but it is likely that what Stewart is claiming is (I guess) that a limit that goes to $-\infty$ or to $+\infty$ on only one side (as in this example, where the limit is only from the left), is "non existing". Otherwise, the statement "the limit does not exist becuase the denomiator approaches $0$ while the numerator does not" would be simply nonsense.